我的代碼非常混亂,但我試圖在我的網站上創建一個論壇。獲取錯誤:類mysqli_result的對象無法轉換爲字符串
<?php
$conn = mysqli_connect("localhost", "root", "");
if($conn->connect_error){
die("Connection Failed");
}
mysqli_select_db($conn, 'forum');
$get= "select * from forum";
$runq = mysqli_query($conn, $get);
while($fetch_value=mysqli_fetch_array($runq)){
$get_usr_name=$fetch_value['username'];
$get_body = $fetch_value['info'];
}
$result = $conn->query($get);
if(is_null($get_body)){
}
else{
echo "<div id='inputform1'>";
for($row = 1; $row < $result->num_rows; $row++){
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "forum";
$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
if(mysqli_connect_errno()){
die("database connection failed: ") .
mysqli_connect_errno() .
" (" . mysqli_connect_errno() . ")";
}
$result1 = mysqli_query($connection, "select info from forum where id='$row';");
$result = mysqli_query($connection, "select username from forum where id='$row';");
echo "<strong class=\"after_username\">by: {$result}</strong><br><br><p id=\"bodyforum\">{$result1}<br><br></p>";
}
echo "</div>";
}
?>
我似乎無法發佈用戶名和密碼;它一直說:
Object of class mysqli_result could not be converted to string
我已經嘗試了許多不同的「戰術」,但沒有一個似乎工作。有人能幫我嗎?
哪一行會拋出錯誤?爲什麼運行相同的查詢兩次? – chris85
chris85是正確的:您可以使用'select info,username from forum where id ='$ id'「'來避免兩次查詢,但是應該從查詢結果中獲取'$ id';假設所有的ID爲你的查詢返回行將返回從1開始,並定期增加。這可能不是這樣的。 – kungphu